LeetCode——Symmetric Tree

# 101

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1/ \
  2   2/ \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1/ \
  2   2\   \
   3    3Note:
Bonus points if you could solve it both recursively and iteratively.

这题是要求判断二叉树是否对称。要求用递归和迭代。这题的思路其实并不难，因为是二叉树。先写出以下几种情况：

1. root为空，返回true,开始递归2. root非空，无子节点，返回true3. 如果不是左右子节点都存在且相等，返回false4. 左右子节点存在且相当，再分别比较左右子树，形成递归

需要第一一个递归函数。这题的递归不难，只要把思路理清，很容易就写出来了。

C++/递归方法

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:bool isSymmetric(TreeNode* root) {if(!root) return true;return isSymmetric(root -> left,root -> right);}bool isSymmetric(TreeNode* left,TreeNode* right){if(!left && !right)return true;if((left && !right) || (!left && right) || (left -> val != right -> val))return false;return isSymmetric(left -> left,right -> right) && isSymmetric(left -> right,right -> left);}
};

迭代的方法其实核心思想是一样的，只需要转换一下就可以了。

C++/迭代方法

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:bool isSymmetric(TreeNode* root) {if(!root)return true;queue<TreeNode*>q1,q2;q1.push(root -> left);q2.push(root -> right);while (!q1.empty() && !q2.empty()){TreeNode* left = q1.front();q1.pop();TreeNode* right = q2.front();q2.pop();if (left == NULL && right == NULL)continue;if (left == NULL || right == NULL)return false;if (left -> val != right -> val)return false;q1.push(left ->l eft);q1.push(left -> right);q2.push(right -> right);q2.push(right -> left);}return true;}
};

使用的是DFS深度优先搜索的算法，其实也可以用BFS广度优先搜索，差别不是很大，不同的比较方式，下面用Python实现。

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def isSymmetric(self, root):""":type root: TreeNode:rtype: bool"""if not root:return Truequeuel, queuer = [root.left], [root.right]while len(queuel) > 0 and len(queuer) > 0:left = queuel.pop()right = queuer.pop()if not left and not right:continueelif not left or not right:return Falseif left.val != right.val:return Falsequeuel.insert(0, left.left)queuel.insert(0, left.right)queuer.insert(0, right.right)queuer.insert(0, right.left)return len(queuel) == 0 and len(queuer) == 0

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LeetCode——Symmetric Tree

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